Solve the following systems of equations:
x − y + z = 4
x − 2y − 2z = 9
2x + y + 3z = 1
Solution
We have,
x − y + z = 4 .....(i)
x − 2y − 2z = 9 ....(ii)
2x + y + 3z = 1 .....(iii)
From equation (i), we get
z = 4 - x + y
=> z = -x + y + 4
Subtracting the value of z in equation (ii), we get
x - 2y - 2(-x + y + 4) = 9
`=> x - 2y + 2x - 2y - 8 = 8`
=> 3x - 4x = 9 + 8
=> 3x - 4y = 17 ......(iv)
Subtracting the value of z in equation (iii), we get
2x + y + 3(-x + y + 4) = 1
=> 2x + y + 3x + 3y + 12 = 1
=> -x + 4y =1 - 12
=> -x + 4y = -11 .....(v)
Adding equations (iv) and (v), we get
3x - x - 4y + 4y = 17 - 11
=> 2x = 6
=> x = 6/2 = 3
Putting x = 3 in equation (iv), we get
`3 xx 2 - 4y = 17`
=> 9 - 4y = 17
=> -4y = 17 - 9
=> -4y = 8
`=> y = 8/(-4) = -2`
Putting x = 3 and y = -2 in z = -x + y + 4 we get
z = -3 -2 + 4
=> x = -5 + 4
=> z= -1
Hence, solution of the giving system of equation is x = 3, y = -2, z = -1